How do I show that the sum of the reciprocals of divisors of a perfect number is $2$?
I tried $d_i\mid n$ with $i\in\mathbb{N},\;d_i\leq n$ then $$\frac{1}{d_1}+\frac{1}{d_2}+\frac{1}{d_3}+…+\frac{1}{d_i}=2$$
$$\sum_{d\mid n} \frac{1}{d}=2$$
So actually, I have to show this last equality, whereas $n$ is a perfect number.
Best Answer
Hint: multiply the first equation by $n$, then note that if the divisors are sorted in increasing order $\frac n{d_1}=?$